eulerian numbers for negative argmuents

[Gottfried Helms]

Am 14.03.2006 15:38 schrieb Mitch Harris:
> > There are 4 combinatorial counting functions that have very similar
> > recurrences
> >
> > binomial:     b(n, m) =        b(n-1, m) +        b(n-1, m-1)
> > stirling 1:  s1(n, m) = (n-1) s1(n-1, m) +       s1(n-1, m-1)
> > stirling 2:  s2(n, m) =     m s2(n-1, m) +       s2(n-1, m-1)
> > eulerian:     e(n, m) = (m+1)  e(n-1, m) + (n-m)  e(n-1, m-1)
> > (with appropriate base cases)
> >
> > counting respectively subsets of [n] of size m, permutations of [n] with
> > m cycles, set partitions of [n] into m nonempty subsets, and
> > permutations of [n] with m ascents.
> >
> > By not caring about the combinatorics, and rearranging these formulas,
> > one can get formulas to compute them for negative arguments (reciprocity
> > formulas). For example,
> >
> >    b(n-1,m) = b(n,m) - b(n-1,m-1)
> >
> > or
> >
> >    b(-n,m) = b(-(n-1),m) - b(-n,m-1)
> >
> > from which it can then be proved that
> >
> >    b(-n,m) = (-1)^m b(n+m-1,n).
> >
> > Similarly, it can be shown that
> >
> >    s1(-n,-m) = (-1)^m s2(m,n).
> >
> >
> > So now I get to what I am really after - is there any known reduction
> > for the Eulerian numbers e(-n,m) (e(-n,-m) is infeasible because of 0
> > denominators) to something else?
> >
> > Pure computation gives the extended table:
> >
> >   -4    1  49/16 8035/1296
> >   -3    1  17/8  355/108   7715/1728
> >   -2    1  5/4   49/36     205/144   5269/3600 5369/3600
> >   -1    1  1/2   1/3       1/4       1/5       1/6
> >    0    1  0     0         0         0         0
> >    1    1
> >    2    1  1
> >    3    1  4     1
> >    4    1  11    11        1
> >    5    1  26    66        26        1
> >
> > (leaving out values that are too big or 0)
> > Multiplying Eul(-n, m) by (m+1)!^n gets the integral version e*(-n, m):
> >
> >   -4    1   49  8035
> >   -3    1   17  710   61720
> >   -2    1   5   49    820   21076 773136
> >   -1    1   1   2     6     24    120
> >    0    1   0   0     0     0     0
> >
> > where (by inspection)
> >        e*(-1, m) = (m)!
> >        e*(-2, m) = A001819(m+1)  (some central factorial #'s)
> >        e*(-n, 1) = A000337(n)   = (n-1)*2^n + 1
> >
> > and this is as far as I've gotten (no other rows or columns in the OEIS.
> > And for the reciprocal formula:
> >
> >    e(-n, m) = ((m-1+n)e(-n+1,m-1) + e(-n+1,m))/(m+1)
> >
> > Well, I just don't see anything obvious (even after multiplying by
> > (m+1)^n). I don't "get" central factorials (combinatorial or otherwise)
> > but that seems the line to pursue (via Stanley 5.8), and of course the
> > OEIS and superseeker don't find enything for any of the higher rows or
> > columns.
> >
> > Any ideas? Leads? Surely this has been looked at before somewhere?
> >
> > Mitch
Mitch,

 it is a time ago, that I fiddled with that numbers. At
http://mathpages.com/home/kmath012/kmath012.htm or
   ...kmath464... there is also some stuff.

 From one of my q&d -tables I just copy&paste a list, which
shows a relatively simple pattern how to compute the
Euler-numbers for an arbitrary row just by row and col
index. I'm too lazy to make that formula for you... it's
way off my head for a long time.


        Z      e(z,0)              e(z,1)               e(z,2)                                     e(z,3)                              e(z,4)
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
         -4 |  1·1^-4  |    1·2^-4 + 3·1^-4   |    1·3^-4 + 3·2^-4 +  6·1^-4    |      1·4^-4 + 3·3^-4 +  6·2^-4 + 10·1^-4   |    1·5^-4 + 3·4^-4 +  6·3^-4 + 10·2^-4 + 15·1^-4
         -3 |  1·1^-3  |    1·2^-3 + 2·1^-3   |    1·3^-3 + 2·2^-3 +  3·1^-3    |      1·4^-3 + 2·3^-3 +  3·2^-3 +  4·1^-3   |    1·5^-3 + 2·4^-3 +  3·3^-3 +  4·2^-3 +  5·1^-3
         -2 |  1·1^-2  |    1·2^-2 + 1·1^-2   |    1·3^-2 + 1·2^-2 +  1·1^-2    |      1·4^-2 + 1·3^-2 +  1·2^-2 +  1·1^-2   |    1·5^-2 + 1·4^-2 +  1·3^-2 +  1·2^-2 +  1·1^-2
         -1 |  1·1^-1  |    1·2^-1 + 0·1^-1   |    1·3^-1 + 0·2^-1 +  0·1^-1    |      1·4^-1 + 0·3^-1 +  0·2^-1 +  0·1^-1   |    1·5^-1 + 0·4^-1 +  0·3^-1 +  0·2^-1 +  0·1^-1
         0  |  1·1^+0  |    1·2^+0 - 1·1^+0   |    1·3^+0 - 1·2^+0 +  0·1^+0    |      1·4^+0 - 1·3^+0 +  0·2^+0 +  0·1^+0   |    1·5^+0 - 1·4^+0 +  0·3^+0 +  0·2^+0 +  0·1^+0
         1  |  1·1^+1  |    1·2^+1 - 2·1^+1   |    1·3^+1 - 2·2^+1 +  1·1^+1    |      1·4^+1 - 2·3^+1 +  1·2^+1 +  0·1^+1   |    1·5^+1 - 2·4^+1 +  1·3^+1 +  0·2^+1 +  0·1^+1
         2  |  1·1^+2  |    1·2^+2 - 3·1^+2   |    1·3^+2 - 3·2^+2 +  3·1^+2    |      1·4^+2 - 3·3^+2 +  3·2^+2 -  1·1^+2   |    1·5^+2 - 3·4^+2 +  3·3^+2 -  1·2^+2 +  0·1^+2
         3  |  1·1^+3  |    1·2^+3 - 4·1^+3   |    1·3^+3 - 4·2^+3 +  6·1^+3    |      1·4^+3 - 4·3^+3 +  6·2^+3 -  4·1^+3   |    1·5^+3 - 4·4^+3 +  6·3^+3 -  4·2^+3 +  1·1^+3
         4  |  1·1^+4  |    1·2^+4 - 5·1^+4   |    1·3^+4 - 5·2^+4 + 10·1^+4    |      1·4^+4 - 5·3^+4 + 10·2^+4 - 10·1^+4   |    1·5^+4 - 5·4^+4 + 10·3^+4 - 10·2^+4 +  5·1^+4

HTH -

Gottfried Helms